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Let $A\subset C([0,1])$ a closed linear subspace with respect to the usual supremum norm satisfying $A\subset C^1([0,1])$. Is $D\colon A\rightarrow C([0,1]), \ f\rightarrow f'$ continuous iff $A$ is finite dimensional?

If $A$ is finite dimensional $D$ is continuous of course. But is the other implication true at all? Wouldn't something like $A:=\overline{\text{span}\{\sin{\left(t+\frac{1}{n}\right)} \ | \ n\in\mathbb{N}\}}$ be a counterexample?

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  • $\begingroup$ So you need some counterexample of infinite dimensional $A$, or you want to prove that every such infinite dimensional $A$ gives discontinuous $D$? $\endgroup$ – Norbert Jul 31 '12 at 7:02
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    $\begingroup$ Did you write out in detail what continuity of $D$ means? Do you know the Arzelà-Ascoli theorem? What does that tell you about the unit ball of $A$ in the sup-norm if $D$ is continuous? $\endgroup$ – t.b. Jul 31 '12 at 7:05
  • $\begingroup$ Yes sure! Continuity is equivalent to boundedness, i.e. $\lVert D f\lVert\leq M \lVert f\lVert$. Arzela-Ascoli states that a set is relative compact iff it's uniformly bounded and equicontinuos. How can I apply this here? In the set $A$ above I tried to define something infinite dimensional, uniformly bounded and equicontinuous. $\endgroup$ – Julian Jul 31 '12 at 7:09
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    $\begingroup$ Well, the condition on continuity you write tells you that the derivatives of functions in the closed unit ball of $A$ are uniformly bounded, right? The mean value theorem then tells you that ... Finally, what do you know about spaces whose unit ball is compact? $\endgroup$ – t.b. Jul 31 '12 at 7:13
  • $\begingroup$ Yes boundedness is clear. I don't see how to prove equicontinuity. the unit ball is compact iff the space is finite dimensional $\endgroup$ – Julian Jul 31 '12 at 7:19
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From the dicussion above it follows that it is enough to show equicontinuiuty of $\mathrm{Ball}_A(0,1)$ provided $D$ is continuous.

Fix $\varepsilon>0$ and take $\delta=(2\Vert D\Vert)^{-1}\varepsilon$. From mean value theorem it follows that for all $x_1,x_2\in [0,1]$ such that $|x_2-x_1|\leq\delta$ and all $f\in\mathrm{Ball}_A(0,1)$ we have $$ |f(x_2)-f(x_1)|=|f'(\xi)(x_2-x_1)|=|f'(\xi)||x_2-x_1|\leq\Vert f'\Vert_\infty\delta\leq $$ $$ \Vert D\Vert\Vert f\Vert_\infty\delta\leq\Vert D\Vert\cdot 1\cdot \frac{\varepsilon}{2\Vert D\Vert}<\varepsilon $$ This means that $\mathrm{Ball}_A(0,1)$ is equicontinuous.

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  • $\begingroup$ The reason why the intended counterexample (the closed span of shifted sines) doesn't work is that it is not contained in $C^1$. It's a good exercise to find out why. $\endgroup$ – t.b. Jul 31 '12 at 7:43

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